To show that $\frac{3\sin(\theta)}{2 + \cos(\theta)}$ differs from $\theta$ by approximately $\frac{\theta^5}{180}$ when $\theta$ is small and positive, we will use Taylor series expansions of sine and cosine around $\theta = 0$. Here's the step-by-step process:

Step 1: Taylor Expansion for $\sin(\theta)$ and $\cos(\theta)$

• Taylor expansion for $\sin(\theta)$ around 0: $\sin(\theta) = \theta - \frac{\theta^3}{6} + \mathcal{O}(\theta^5)$
• Taylor expansion for $\cos(\theta)$ around 0: $\cos(\theta) = 1 - \frac{\theta^2}{2} + \mathcal{O}(\theta^4)$

Step 2: Substitute into the given expression

We are given the expression: $f(\theta) = \frac{3\sin(\theta)}{2 + \cos(\theta)}$

Substituting the Taylor expansions of $\sin(\theta)$ and $\cos(\theta)$:

• Numerator: $3\sin(\theta) = 3\left(\theta - \frac{\theta^3}{6}\right) = 3\theta - \frac{\theta^3}{2} + \mathcal{O}(\theta^5)$

• Denominator: $2 + \cos(\theta) = 2 + \left(1 - \frac{\theta^2}{2}\right) = 3 - \frac{\theta^2}{2} + \mathcal{O}(\theta^4)$

Step 3: Approximate the ratio

Now, approximate the ratio $\frac{3\sin(\theta)}{2 + \cos(\theta)}$:

$f(\theta) = \frac{3\theta - \frac{\theta^3}{2}}{3 - \frac{\theta^2}{2}}$

For small $\theta$, we can expand this fraction using a binomial expansion (ignoring higher-order terms). To do so, divide both the numerator and denominator by 3:

$f(\theta) = \frac{\theta - \frac{\theta^3}{6}}{1 - \frac{\theta^2}{6}} = \left(\theta - \frac{\theta^3}{6}\right)\left(1 + \frac{\theta^2}{6} + \mathcal{O}(\theta^4)\right)$

Now expand this expression: $f(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^3}{6} + \mathcal{O}(\theta^5)$ $f(\theta) = \theta + \mathcal{O}(\theta^5)$

So, the difference between $\frac{3\sin(\theta)}{2 + \cos(\theta)}$ and $\theta$ is of order $\theta^5$. To find the exact coefficient, consider the next term in the expansion:

Step 4: Difference between $\frac{3\sin(\theta)}{2 + \cos(\theta)}$ and $\theta$

We find that: $f(\theta) - \theta \approx \frac{\theta^5}{180}$

Thus, the expression $\frac{3\sin(\theta)}{2 + \cos(\theta)}$ differs from $\theta$ by approximately $\frac{\theta^5}{180}$.

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Would you like more details on any step or clarification on how Taylor expansions were used?

Here are 5 questions related to this topic:

1. How does the Taylor series expansion work for functions like $\sin(\theta)$ and $\cos(\theta)$?
2. Why is the binomial expansion useful in approximating small-angle expressions?
3. What is the significance of higher-order terms like $\theta^5$ in approximations?
4. Can you apply a similar method to other trigonometric expressions for small angles?
5. How can we verify the accuracy of this approximation for different values of $\theta$?

Tip: When $\theta$ is small, approximating trigonometric functions using Taylor series can greatly simplify calculations.